Problem: The lifespans of seals in a particular zoo are normally distributed. The average seal lives $17.9$ years; the standard deviation is $3.6$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a seal living longer than $7.1$ years.
Answer: $17.9$ $14.3$ $21.5$ $10.7$ $25.1$ $7.1$ $28.7$ $99.7\%$ $0.15\%$ $0.15\%$ We know the lifespans are normally distributed with an average lifespan of $17.9$ years. We know the standard deviation is $3.6$ years, so one standard deviation below the mean is $14.3$ years and one standard deviation above the mean is $21.5$ years. Two standard deviations below the mean is $10.7$ years and two standard deviations above the mean is $25.1$ years. Three standard deviations below the mean is $7.1$ years and three standard deviations above the mean is $28.7$ years. We are interested in the probability of a seal living longer than $7.1$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the seals will have lifespans within 3 standard deviations of the average lifespan. The remaining $0.3\%$ of the seals will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({0.15\%})$ will live less than $7.1$ years and the other half $({0.15\%})$ will live longer than $28.7$ years. The probability of a particular seal living longer than $7.1$ years is ${99.7\%} + {0.15\%}$, or $99.85\%$.